We consider the system Applicative_05__TakeDropWhile.

  Alphabet:

    cons : [a * c] --> c 
    dropWhile : [a -> b * c] --> c 
    if : [b * c * c] --> c 
    nil : [] --> c 
    takeWhile : [a -> b * c] --> c 
    true : [] --> b 

  Rules:

    if(true, x, y) => x 
    if(true, x, y) => y 
    takeWhile(f, nil) => nil 
    takeWhile(f, cons(x, y)) => if(f x, cons(x, takeWhile(f, y)), nil) 
    dropWhile(f, nil) => nil 
    dropWhile(f, cons(x, y)) => if(f x, dropWhile(f, y), cons(x, y)) 

This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0).

We use the dependency pair framework as described in [Kop12, Ch. 6/7], with static dependency pairs (see [KusIsoSakBla09] and the adaptation for AFSMs and accessible arguments in [Kop13]).

We thus obtain the following dependency pair problem (P_0, R_0, static, formative):

  Dependency Pairs P_0:

    0] takeWhile#(F, cons(X, Y)) =#> if#(F X, cons(X, takeWhile(F, Y)), nil)   
    1] takeWhile#(F, cons(X, Y)) =#> takeWhile#(F, Y)   
    2] dropWhile#(F, cons(X, Y)) =#> if#(F X, dropWhile(F, Y), cons(X, Y))   
    3] dropWhile#(F, cons(X, Y)) =#> dropWhile#(F, Y)   

  Rules R_0:

    if(true, X, Y) => X 
    if(true, X, Y) => Y 
    takeWhile(F, nil) => nil 
    takeWhile(F, cons(X, Y)) => if(F X, cons(X, takeWhile(F, Y)), nil) 
    dropWhile(F, nil) => nil 
    dropWhile(F, cons(X, Y)) => if(F X, dropWhile(F, Y), cons(X, Y)) 

Thus, the original system is terminating if (P_0, R_0, static, formative) is finite.

We consider the dependency pair problem (P_0, R_0, static, formative).

We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows:

    * 0 :  
    * 1 : 0, 1 
    * 2 :  
    * 3 : 2, 3 

This graph has the following strongly connected components:

  P_1:

    takeWhile#(F, cons(X, Y)) =#> takeWhile#(F, Y)   

  P_2:

    dropWhile#(F, cons(X, Y)) =#> dropWhile#(F, Y)   

By [Kop12, Thm. 7.31], we may replace any dependency pair problem (P_0, R_0, m, f) by (P_1, R_0, m, f) and (P_2, R_0, m, f).

Thus, the original system is terminating if each of (P_1, R_0, static, formative) and (P_2, R_0, static, formative) is finite.

We consider the dependency pair problem (P_2, R_0, static, formative).

We will use the reduction pair processor [Kop12, Thm. 7.16].  It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

  dropWhile#(F, cons(X, Y)) >? dropWhile#(F, Y) 
  if(true, X, Y) >= X 
  if(true, X, Y) >= Y 
  takeWhile(F, nil) >= nil 
  takeWhile(F, cons(X, Y)) >= if(F X, cons(X, takeWhile(F, Y)), nil) 
  dropWhile(F, nil) >= nil 
  dropWhile(F, cons(X, Y)) >= if(F X, dropWhile(F, Y), cons(X, Y)) 

We orient these requirements with a polynomial interpretation in the natural numbers.

The following interpretation satisfies the requirements:

  cons = \y0y1.3 + 2y1 
  dropWhile = \G0y1.3 + 2y1 
  dropWhile# = \G0y1.y1 
  if = \y0y1y2.y1 + y2 
  nil = 0 
  takeWhile = \G0y1.3y1 
  true = 0 

Using this interpretation, the requirements translate to:

  [[dropWhile#(_F0, cons(_x1, _x2))]] = 3 + 2x2 > x2 = [[dropWhile#(_F0, _x2)]] 
  [[if(true, _x0, _x1)]] = x0 + x1 >= x0 = [[_x0]] 
  [[if(true, _x0, _x1)]] = x0 + x1 >= x1 = [[_x1]] 
  [[takeWhile(_F0, nil)]] = 0 >= 0 = [[nil]] 
  [[takeWhile(_F0, cons(_x1, _x2))]] = 9 + 6x2 >= 3 + 6x2 = [[if(_F0 _x1, cons(_x1, takeWhile(_F0, _x2)), nil)]] 
  [[dropWhile(_F0, nil)]] = 3 >= 0 = [[nil]] 
  [[dropWhile(_F0, cons(_x1, _x2))]] = 9 + 4x2 >= 6 + 4x2 = [[if(_F0 _x1, dropWhile(_F0, _x2), cons(_x1, _x2))]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_2, R_0) by ({}, R_0).  By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed.

Thus, the original system is terminating if (P_1, R_0, static, formative) is finite.

We consider the dependency pair problem (P_1, R_0, static, formative).

We will use the reduction pair processor [Kop12, Thm. 7.16].  It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

  takeWhile#(F, cons(X, Y)) >? takeWhile#(F, Y) 
  if(true, X, Y) >= X 
  if(true, X, Y) >= Y 
  takeWhile(F, nil) >= nil 
  takeWhile(F, cons(X, Y)) >= if(F X, cons(X, takeWhile(F, Y)), nil) 
  dropWhile(F, nil) >= nil 
  dropWhile(F, cons(X, Y)) >= if(F X, dropWhile(F, Y), cons(X, Y)) 

We orient these requirements with a polynomial interpretation in the natural numbers.

The following interpretation satisfies the requirements:

  cons = \y0y1.1 + 2y1 
  dropWhile = \G0y1.3 + 3y1 
  if = \y0y1y2.y1 + y2 
  nil = 0 
  takeWhile = \G0y1.y1 
  takeWhile# = \G0y1.y1 
  true = 0 

Using this interpretation, the requirements translate to:

  [[takeWhile#(_F0, cons(_x1, _x2))]] = 1 + 2x2 > x2 = [[takeWhile#(_F0, _x2)]] 
  [[if(true, _x0, _x1)]] = x0 + x1 >= x0 = [[_x0]] 
  [[if(true, _x0, _x1)]] = x0 + x1 >= x1 = [[_x1]] 
  [[takeWhile(_F0, nil)]] = 0 >= 0 = [[nil]] 
  [[takeWhile(_F0, cons(_x1, _x2))]] = 1 + 2x2 >= 1 + 2x2 = [[if(_F0 _x1, cons(_x1, takeWhile(_F0, _x2)), nil)]] 
  [[dropWhile(_F0, nil)]] = 3 >= 0 = [[nil]] 
  [[dropWhile(_F0, cons(_x1, _x2))]] = 6 + 6x2 >= 4 + 5x2 = [[if(_F0 _x1, dropWhile(_F0, _x2), cons(_x1, _x2))]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_1, R_0) by ({}, R_0).  By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed.

As all dependency pair problems were succesfully simplified with sound (and complete) processors until nothing remained, we conclude termination.


+++ Citations +++

[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.
[Kop13]  C. Kop.  Static Dependency Pairs with Accessibility.  Unpublished manuscript, http://cl-informatik.uibk.ac.at/users/kop/static.pdf, 2013.
[KusIsoSakBla09]  K. Kusakari, Y. Isogai, M. Sakai, and F. Blanqui.  Static Dependency Pair Method Based On Strong Computability for Higher-Order Rewrite Systems.  In volume 92(10) of IEICE Transactions on Information and Systems.  2007--2015, 2009.